How do you integrate #(x^2e^(x/2))dx#?
1 Answer
Jan 23, 2017
Explanation:
#I=intx^2e^(x/2)dx#
Use the substitution
#{(2t=x),(2dt=dx),(4t^2=x^2):}#
Then:
#I=intx^2e^(x/2)dx=int(4t^2)(e^t)(2dt)=int8t^2e^tdt#
Now use integration by parts. This takes the form
#{(u=8t^2" "=>" "du=16tdt),(dv=e^tdt" "=>" "v=e^t):}#
So:
#I=uv-intvdu=8t^2e^t-int16te^tdt#
Again perform integration by parts with a knew
#{(u=16t" "=>" "du=16dt),(dv=e^tdt" "=>" "v=e^t):}#
Hence:
#I=8t^2e^t-(uv-intvdu)#
#I=8t^2e^t-16te^t+int16e^tdt#
Note that
#I=8t^2e^t-16te^t+16e^t+C#
#I=8e^t(t^2-2t+2)+C#
Returning to
#I=8e^(x/2)(x^2/4-x+2)+C#
#I=2e^(x/2)(x^2-4x+8)+C#